Trig
posted by Lisa on .
Find all solutions in the interval 0 degrees<θ<360 degrees. If rounding necessary, round to the nearest tenth of a degree. 17sec2 θ − 15tanθsecθ − 15 = 0

17sec^2θ  15secθ tanθ  15 = 0
15secθ tanθ = 17sec^2θ  15
225 sec^2θ tan^2θ = 289sec^4θ  510sec^2θ + 225
225sec^4θ  225sec^2θ = 289sec^4θ  510sec^2θ + 225
64sec^4θ  285sec^2θ + 225 = 0
That's just a quadratic in sec^2θ, so just solve it and you have your solution candidates.
However, because we squared things, there may be spurious solutions, so you have to check the values in the original equation.
In radians, I get .159,1.000,2.141,2.982 
17sec2 θ − 15tanθsecθ − 15 = 0
17/cos^2 Ø  15(sinØ/cosØ)(1/cosØ)  15 = 0
times cos^2 Ø
17  15sinØ  15cos^2 Ø = 0
17  15sinØ  15(1  sin^2 Ø) = 0
15sin^2 Ø  15sinØ + 2 = 0
sinØ = (15 ± √105)/30
sinØ = .158435 or sinØ = .841565
Ø = 9.1° or 170.9° or Ø = 57.3° or 122.7°